A projectile of mass 1.100 kg is shot straight up with an initial speed of 29.0 m/s.

In summary, a projectile with a mass of 1.100 kg is shot straight up at an initial speed of 29.0 m/s. The question asks for the maximum height it would reach without air friction, and the magnitude of the average force due to air resistance if it reaches a maximum height of 35.2 m. The second law of Newton and the kinematic equations can be used to find the acceleration and time, which can then be used to calculate the force due to air resistance.
  • #1
IShouldBSurfing
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Hello ! Wow, you guys are awesome, Thanks for all the help ! I've got another one I'm stuck on - ... I got part A.. the answer is 42.9082, but I can't get part B... ? help pleaaseee? thaaanks

A projectile of mass 1.100 kg is shot straight up with an initial speed of 29.0 m/s.
(a) How high would it go if there were no air friction?
m
(b) If the projectile rises to a maximum height of only 35.2 m, determine the magnitude of the average force due to air resistance.
N
 
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  • #2
Use the second law of Newton: You would have -mg - Fr = ma

similarly, by the kinematic equations, you can find aceleration by using a system of equations: v = v0 + at and x = x0 + v0t + (1/2)at^2

in the first, v is o since at the top the velocity is o, you really have two equations with two uknownsl; t and a.

once you have "a" you can use -mg - Fr = ma or Fr = mg-ma

Hope i was some help here. (tell me if i did something wrong, i make mistakes sometimes)
 
  • #3
hm.. i didn't really understand what u said at all.. how am i supposed to find a when i don't know t ? ah, confused
 
  • #4
its a system of equations that gets you t and a. The systems are
-29 = at
35.2 = 29t + (1/2)at^2
these two help you solve for a AND t (not that you need t)
does that help?
 

1. How high will the projectile go?

The maximum height reached by the projectile can be calculated using the formula h = (v0^2) / (2g), where h is the maximum height, v0 is the initial velocity, and g is the acceleration due to gravity (9.8 m/s^2). In this case, the maximum height would be approximately 43.4 meters.

2. How long will it take for the projectile to reach its highest point?

The time it takes for the projectile to reach its highest point can be calculated using the formula t = v0 / g, where t is the time and v0 is the initial velocity. In this case, the projectile would take approximately 2.96 seconds to reach its highest point.

3. What is the velocity of the projectile at its highest point?

At the highest point, the velocity of the projectile is equal to 0 m/s. This is because the projectile stops at its highest point before falling back down due to the force of gravity.

4. How far will the projectile travel horizontally before hitting the ground?

The horizontal distance traveled by the projectile can be calculated using the formula d = v0 * t, where d is the distance, v0 is the initial velocity, and t is the time. In this case, the projectile would travel approximately 85.6 meters before hitting the ground.

5. What is the velocity of the projectile when it hits the ground?

The final velocity of the projectile when it hits the ground is equal to its initial velocity (29.0 m/s) multiplied by the coefficient of restitution (e), which represents the ratio of final velocity to initial velocity after a collision. In most cases, e is less than 1, meaning the final velocity will be less than the initial velocity. However, if we assume an ideal scenario where e = 1, the final velocity would also be 29.0 m/s.

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